Problem 118 - Chemistry: The Central Science in SI Units, Global Edition

Question

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathbf{a})\) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at 99.7 \(\mathrm{kPa}\) and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

Step-by-Step Solution

Answer

The molecular formula of cyclopropane is C3H6. It is likely to deviate more from ideal gas behavior at moderately high pressures and room temperature compared to argon due to its more complex structure and stronger intermolecular forces. Cyclopropane would effuse more slowly than methane as its effusion rate is approximately 0.62 times that of methane.

Step 1: Determine the empirical formula

Given the percentage composition by mass as \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\), we can find the moles of each element by assuming we have a 100 g sample of the gas. \(85.7 \, \mathrm{g \, C * \frac{1 \, mol \, C}{12.01 \, g \, C}} \approx 7.14 \, \mathrm{mol \, C}\) \(14.3 \, \mathrm{g \, H * \frac{1 \, mol \, H}{1.008 \, g \, H}} \approx 14.2 \, \mathrm{mol \, H}\) Now, divide the moles of both elements by the minimum value. \(\frac{7.14 \, \mathrm{mol \, C}}{7.14} \approx 1 \, \mathrm{mol \, C}\) \(\frac{14.2 \, \mathrm{mol \, H}}{7.14} \approx 2 \, \mathrm{mol \, H}\) The empirical formula of cyclopropane is CH2.

Step 2: Calculate the molar mass

We are given a sample of the gas with mass 1.56 g and volume 1.00 L under the conditions 99.7 kPa and 50.0°C. Use the Ideal Gas Law to determine the molar mass of the gas. \(PV = nRT \) \[n = \frac{PV}{RT} \] Here, \(P = 99.7 \, \mathrm{kPa}\), \(R = 8.314 \, \mathrm{\frac{J}{mol \cdot K}}\), \(T = 50.0 + 273.15 = 323.15 \, \mathrm{K}\), and \(V = 1.00 \, \mathrm{L} = 1000 \, \mathrm{mL}\). Convert pressure from kPa to Pa. \(P = 99.7 \, \mathrm{kPa} * 1000 \, \mathrm{\frac{Pa}{kPa}} = 99700 \, \mathrm{Pa}\) Now, plug in the values and calculate the number of moles: \(n = \frac{(99700 \, \mathrm{Pa})(1.00 \times 10^{-3} \, \mathrm{m^3})}{(8.314 \, \mathrm{\frac{J}{mol \cdot K}})(323.15 \, \mathrm{K})} \approx 0.0373 \, \mathrm{mol}\) Next, find the molar mass by dividing the mass by the moles: \(Molar \, Mass = \frac{1.56 \, \mathrm{g}}{0.0373 \, \mathrm{mol}} \approx 41.8 \, \mathrm{\frac{g}{mol}}\)

Step 3: Determine the molecular formula

The empirical formula of cyclopropane is CH2, which has a molar mass of approximately 14.02 g/mol. Divide the molar mass of the compound by the empirical formula's molar mass and round to the nearest whole number: \(\frac{41.8 \, \mathrm{\frac{g}{mol}}}{14.02 \, \mathrm{\frac{g}{mol}}} \approx 3\) The molecular formula of cyclopropane is C3H6, which is obtained by multiplying the empirical formula by the factor 3.

Step 4: Deviation from ideal-gas behavior

Cyclopropane's molecular formula, C3H6, suggests that it is a more complex molecule than a noble gas like argon. As a result, it is more likely to have stronger intermolecular forces and therefore deviate more from ideal gas behavior, particularly at moderately high pressures and room temperature.

Step 5: Effusion comparison with methane

The effusion rate of a gas is inversely proportional to the square root of its molar mass. To compare the effusion rates of cyclopropane and methane, we can use the following relation: \(\frac{Rate_{C3H6}}{Rate_{CH4}} = \sqrt{\frac{Molar \, Mass_{CH4}}{Molar \, Mass_{C3H6}}}\) \(Molar \, Mass_{CH4} = 16.04 \, \mathrm{\frac{g}{mol}}\) \(\frac{Rate_{C3H6}}{Rate_{CH4}} = \sqrt{\frac{16.04 \, \mathrm{\frac{g}{mol}}}{41.8 \, \mathrm{\frac{g}{mol}}}} \approx 0.62\) Since the rate is less than 1, cyclopropane would effuse more slowly than methane.

Key Concepts

Empirical Formula

The empirical formula is a simplified representation of a compound that shows the smallest whole-number ratio of elements present. To find the empirical formula, it's crucial to start with the percentage by mass of each element in the compound. Let's use cyclopropane as an example. If it contains 85.7% carbon (\(\mathrm{C}\)) and 14.3% hydrogen (\(\mathrm{H}\)), we can assume a sample size of 100 grams. This means we have 85.7 grams of carbon and 14.3 grams of hydrogen.

Next, convert these masses to moles using their respective atomic masses. For carbon, the atomic mass is approximately 12.01 g/mol, and for hydrogen, it is 1.008 g/mol. Calculate the moles of carbon and hydrogen:

• Carbon: \(\frac{85.7 \, \mathrm{g}}{12.01 \, \mathrm{g/mol}} \approx 7.14 \, \mathrm{mol\, C}\)
• Hydrogen: \(\frac{14.3 \, \mathrm{g}}{1.008 \, \mathrm{g/mol}} \approx 14.2 \, \mathrm{mol\, H}\)

To find the empirical formula, divide by the smallest number of moles to get whole numbers, leading to CH2 as the empirical formula of cyclopropane.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that relates the pressure, volume, temperature, and amount of gas. It's expressed as \(PV = nRT\), where:

• \(P\) is the pressure of the gas, measured in pascals (Pa).
• \(V\) is the volume of the gas, often in liters (L).
• \(n\) is the number of moles of the gas.
• \(R\) is the ideal gas constant (\(8.314 \, \mathrm{J/(mol \cdot K)}\)).
• \(T\) is the temperature in Kelvin (K).

The Ideal Gas Law is useful to calculate a gas's molar mass when its mass, volume, pressure, and temperature are known. In cyclopropane's case, we have a mass of 1.56 g, a volume of 1.00 L, a pressure of 99.7 kPa (converted to 99,700 Pa), and a temperature of 50°C (converted to 323.15 K). By rearranging the equation to find \(n\), we can solve it to get the number of moles and further determine the molar mass of the gas, leading to its molecular formula.

Effusion Rate

Effusion is the process by which gas molecules escape through a small hole into a vacuum. Graham's law of effusion states that the effusion rate of a gas is inversely proportional to the square root of its molar mass. This can be mathematically expressed as:

• \(\frac{\text{Rate of Effusion of Gas A}}{\text{Rate of Effusion of Gas B}} = \sqrt{\frac{\text{Molar Mass of Gas B}}{\text{Molar Mass of Gas A}}}\)

Applying this to cyclopropane (\(\mathrm{C_3H_6}\)) and methane (\(\mathrm{CH_4}\)), with molar masses approximately 41.8 g/mol and 16.04 g/mol respectively, cyclopropane will effuse more slowly than methane because its higher molar mass results in a smaller effusion rate. This principle allows us to compare the behavior of different gases under the same conditions.

Cyclopropane

Cyclopropane is a small cyclic hydrocarbon with the molecular formula \(\mathrm{C_3H_6}\). This gas is notable for its use in anesthetics due to its potency when inhaled with oxygen. Unlike linear hydrocarbons, cyclopropane's ring structure imparts unique chemical properties. It is more reactive than its linear counterparts because of the strained bonds in its triangular configuration.

It is also important when discussing gas behaviors, such as deviation from the ideal gas law. Cyclopropane, because of its fairly complex molecular structure compared to a monoatomic gas like argon, may experience stronger intermolecular forces. This tends to cause a deviation from ideal behavior, especially under conditions of higher pressure and room temperature. Understanding these properties and applying this knowledge can aid in better predictions of its behavior in practical applications.